the sides of a triangle have lengths 4x+1, 2x+1 and 6x-1. if the lengths of the longest side is 6x-1, what values of x make the triangle obtuse?
Well, to determine if a triangle is obtuse, we need to use the Pythagorean Theorem. In an obtuse triangle, the square of the longest side is greater than the sum of the squares of the other two sides.
Using the Pythagorean Theorem, we have:
(6x - 1)^2 > (4x + 1)^2 + (2x + 1)^2
Expanding this expression, we get:
36x^2 - 12x + 1 > 16x^2 + 8x + 1 + 4x^2 + 4x + 1
Simplifying further, we have:
16x^2 - 24x > 0
Dividing both sides by 8, we get:
2x^2 - 3x > 0
Now, let's factorize:
x(2x - 3) > 0
We have two cases:
1) x > 0 and 2x - 3 > 0
2) x < 0 and 2x - 3 < 0
For case 1, solving x > 0 and 2x - 3 > 0, we get:
x > 0 and x > 3/2
Since both conditions need to be satisfied, x > 3/2.
For case 2, solving x < 0 and 2x - 3 < 0, we get:
x < 0 and x < 3/2
Again, both conditions need to be satisfied, so x < 0.
Therefore, the values of x that make the triangle obtuse are x < 0 and x > 3/2.
To determine the values of x that make the triangle obtuse, we need to use the triangle inequality theorem and the Pythagorean theorem.
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Let's denote the lengths of the sides as follows:
Side 1: 4x + 1
Side 2: 2x + 1
Side 3 (Longest side): 6x - 1
For the triangle to be obtuse, the square of the longest side (Side 3) must be greater than the sum of the squares of the other two sides (Side 1 and Side 2) according to the Pythagorean theorem.
So, we have the following inequality to solve:
(6x - 1)^2 > (4x + 1)^2 + (2x + 1)^2
Expanding the equation:
36x^2 - 12x + 1 > 16x^2 + 8x + 1 + 4x^2 + 4x + 1
Combining like terms:
36x^2 - 12x + 1 > 24x^2 + 12x + 2
Subtracting 24x^2, 12x, and 2 from both sides:
12x^2 - 24x - 1 > 0
Now, let's solve for x in this quadratic inequality. We can use either factoring or the quadratic formula.
Using factoring is difficult in this case, so let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
a = 12, b = -24, c = -1
x = (-(-24) ± sqrt((-24)^2 - 4(12)(-1))) / (2(12))
x = (24 ± sqrt(576 + 48)) / 24
x = (24 ± sqrt(624)) / 24
x = (24 ± sqrt(16 * 39)) / 24
x = (24 ± 4√39) / 24
x = (6 ± √39) / 6
So, the values of x that make the triangle obtuse are:
x = (6 + √39) / 6 or x = (6 - √39) / 6
To determine when a triangle is obtuse, we need to use the Triangle Inequality Theorem and the definition of an obtuse triangle.
The Triangle Inequality Theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.
Let's apply this to the given triangle with side lengths 4x+1, 2x+1, and 6x-1:
The longest side is 6x-1. According to the Triangle Inequality Theorem, the sum of the other two sides must be greater than the length of the longest side.
So we have two inequalities:
(1) 4x+1 + 2x+1 > 6x-1
(2) 4x+1 + 6x-1 > 2x+1
Let's solve these inequalities to find the range of values for x that will make the triangle obtuse.
(1) 6x + 2 > 6x - 1
2 > -1
Since 2 is always greater than -1, the first inequality is true for all values of x.
(2) 4x + 6x > 2x + 1 - 1
10x > 2x
Subtracting 2x from both sides, we get:
8x > 0
Dividing both sides by 8 (since we want to solve for x), we get:
x > 0
Therefore, the values of x that make the triangle obtuse are any positive values of x.
Summary: For any positive value of x, the triangle with side lengths 4x+1, 2x+1, and 6x-1 will be an obtuse triangle.
(6x-1)^2 > (4x+1)^2 + (2x+1)^2 ,
36x^2 - 12x + 1 = 16x^2 + 8x + 1 + 4x^2 + 4x + 1
6x^2 - 24x -1 > 0
suppose we solve the corresponding equation
6x^2 - 24x -1 = 0
x = (24 ± √600)/12
=(24 ± 10√6)/24
= (12 ± 5√6)/12 = appr 2.0206 or a negative
but clearly 6x-1 > 0
x > 1/6
so it is obtuse for x > 2.0206